9702_m22_qp_42-neha

Question

(a) The point P in Fig. 1.1 represents a point mass.

On Fig. 1.1, draw lines to represent the gravitational field around P.

Fig. 1.1

(b) A moon is in circular orbit around a planet.

Explain why the path of the moon is circular.

The angular velocity of a moon is ω when the orbit of the moon has a radius r about the planet.

Fig. 1.2 shows the variation of \(r^{3}\) with \(1/ω^{2}\) for these moons.

(i) Show that the mass M of the planet is given by the expression

\(M = \frac{gradient}{G}\)

where G is the gravitational constant.

(ii) Use Fig. 1.2 and the expression in (c)(i) to show that the mass M of the planet is \(1.0 \times 10^{26}\) kg.

(iii) Determine the speed of a moon in orbit around the planet with an orbital radius of \(1.2 \times 10^{8}\)m.

Answer/Explanation

Ans:

(a) at least \(4\) straight radial lines to P

all arrows pointing along the lines towards P

(b) Any 2 from:

gravitational force provides the centripetal force

(centripetal or gravitational) force has constant magnitude

(centripetal or gravitational) force is perpendicular to velocity (of moon) / direction of motion (of moon)

\(M\) = \(\frac{r^{3}\omega ^{2}}{G}\) and gradient = \(r^{2}\omega ^{2}\) hence M = \(\frac{gradient}{G}\)

\(r^{3}\) = \(GM \times 1/\omega ^{2}\) so gradient = GM hence M = \(\frac{gradient}{G}\)

(ii) \(M = 4.1\times 10^{23}/(6.0 \times 10^{7}\times 6.67\times 10^{-11})\) = \(1.0 \times 10^{26}\) kg

(iii) \(\frac{GMm}{r^{2}}\) = \(\frac{mv^{2}}{r}\)

\(\frac{GM}{r}\) = \(v^{2}\)

\(v^{2}\) = \(\frac{6.67\times 10^{-11}\times 1.0\times 10^{26}}{1.2\times 10^{8}}\)

\(v^{2}\) = \(5.6 \times 10^{7}\, ms^{-1}\)

\(v\) = \(7500\, ms^{-1}\)

Question

A fixed mass of an ideal gas has a volume V and a pressure p. The gas undergoes a cycle of changes, X to Y to Z to X, as shown in Fig. 2.1.

Table 2.1 shows data for p, V and temperature T for the gas at points X, Y and Z.

Table 2.1

(a) State the change in internal energy ΔU for one complete cycle, XYZX.

(b) Calculate the amount n of gas.

Use the space below for any working.

(d) (i) The first law of thermodynamics for a system may be represented by the equation

ΔU = q + W.

State, with reference to the system, what is meant by:

(ii) Explain how the first law of thermodynamics applies to the change Z to X.

Answer/Explanation

Ans:

(a) \(0\)

(b) \(pV\) = \(nRT\)

\((n =) 1.5 \times 10^{5} \times 4.2 \times 10^{–3} / 8.31 \times 540\)

= 0.14 mol

both missing volumes \(1.8 ( \times 10^{–3})\)

(d) (i) \((ΔU:)\) increase in internal energy (of the system)

\((q:)\) thermal energy supplied to the system B1

\((W:)\) work done on system

(ii) volume increases and work is done by the gas

temperature decreases and internal energy decreases

Question

A small wooden block (cuboid) of mass m floats in water, as shown in Fig. 3.1.

The top face of the block is horizontal and has area A. The density of the water is ρ.

(a) State the names of the two forces acting on the block when it is stationary.

(b) The block is now displaced downwards as shown in Fig. 3.2 so that the surface of the water is higher up the block.

State and explain the direction of the resultant force acting on the wooden block in this position.

The resultant force F acting on the block is given by

F = –Agρx

where g is the gravitational field strength and x is the vertical displacement of the block from the equilibrium position.

(i) Explain why the oscillations of the block are simple harmonic.

(ii) Show that the angular frequency ω of the oscillations is given by

\(\omega = \sqrt{\frac{A\rho g}{m}}\).

(d) The block is now placed in a liquid with a greater density. The block is displaced and released so that it oscillates vertically. The variation with displacement x of the acceleration a of the block

is measured for the first half oscillation, as shown in Fig. 3.3.

(i) Explain why the maximum negative displacement of the block is not equal to its maximum positive displacement.

(ii) The mass of the block is \(0.57\)kg.

Use Fig. 3.3 to determine the decrease ΔE in energy of the oscillation for the first half oscillation.

Answer/Explanation

Ans:

(a) upthrust, weight

(b) upthrust greater than weight so (resultant force is) upwards

minus sign means F and x are in opposite directions

(ii) \((a = \frac{F}{m}so)\) a = (-)\(\frac{Ag\rho x}{m}\)

so \(\omega ^{2} = \frac{Ag\rho }{m} hence \omega = \sqrt{\frac{Ag\rho }{m}}\)

(d) (i) damping due to viscous forces

(ii) \((E =) 1/2 m\omega ^{2}x_{0}^{2}\)

\(\omega ^{2}\) = (-) gradient

\((E =) 1/2 m\omega ^{2}(x_{1}^{2} – x_{2}^{2})\)

= \(1/2 \times 0.57 \times (2.3/0.020)(0.020^{2} – 0.016^{2})\)

= \(4.7 \times 10^{-3}\) J

Question

(a) State what is represented by an electric field line.

(b) Two point charges P and Q are placed \(0.120\) m apart as shown in Fig. 4.1.

(i) The charge of P is \(+4.0\) nC and the charge of Q is \(–7.2\) nC.

Determine the distance from P of the point on the line joining the two charges where the electric potential is zero.

(ii) State and explain, without calculation, whether the electric field strength is zero at the same point at which the electric potential is zero.

(iii) An electron is positioned at point X, equidistant from both P and Q, as shown in Fig. 4.2.

On Fig. 4.2, draw an arrow to represent the direction of the resultant force acting on the electron.

Answer/Explanation

Ans:

(a) direction of force

force on a positive charge

(b) (i) \(V\) = \(\frac{Q}{4\pi \varepsilon _{0}r}\)

\(\frac{4.0 \times 10^{-9}}{4\pi \varepsilon _{0}x} + \frac{-7.2\times 10^{-9}}{4\pi \varepsilon _{0}(0.120 – x)}\)

\(4(0.120 – x)\) = \(7.2\) x

x = \(0.043\) m

(ii) fields are in the same direction so no

(iii) straight arrow drawn leftwards from X in direction between extended line joining Q and X and the horizontal

Question

The variation with potential difference V of the charge Q on one of the plates of a capacitor is shown in Fig. 5.1.

The capacitor is connected to an 8.0V power supply and two resistors R and S as shown in Fig. 5.2.

The resistance of R is \(25\) kΩ and the resistance of S is \(220\) kΩ.

The switch can be in either position X or position Y.

(a) The switch is in position X so that the capacitor is fully charged.

Calculate the energy E stored in the capacitor.

(b) The switch is now moved to position Y.

(i) Show that the time constant of the discharge circuit is \(3.3\)s.

(ii) The fully charged capacitor in (a) stores energy E.

Determine the time t taken for the stored energy to decrease from E to E/9

State and explain the change, if any, to the time constant of the discharge circuit.

Answer/Explanation

Ans:

(a) (energy stored =) area under line or \(1/2\) QV

= \(1/2 \times 8.0 \times 1.2 \times 10^{-4}\)

= \(4.8 \times 10^{–4}\) J

(b) (i) (τ=) RC

(τ=) \(220 \times 10^{3} \times (1.2 \times 10^{-4}/8.0)\) = \(3.3\) s

(ii) \(E\) ∝ \(V^{2}\)

(so time to) \(V_{0}/3\)

\(V\) = \(V_{0}e^{-t/RC}\)

\(\frac{V_{0}}{3}\) = \(V_{0}e^{-t/3.3}\)

\(\frac{1}{3}\) = \(e^{-t/3.3}\)

\(t\) = \(3.6\) s

time constant is doubled

Question

A small solenoid of area of cross section \(1.6 \times 10^{–3} m^{2}\) is placed inside a larger solenoid of area of cross-section \(6.4 \times 10^{–3}m^{2}\), as shown in Fig. 6.1.

The larger solenoid has \(600\) turns and is attached to a d.c. power supply to create a magnetic field.

The smaller solenoid has \(3000\) turns.

(a) Compare the magnetic flux in the two solenoids.

(b) Compare the magnetic flux linkage in the two solenoids.

(ii) The terminals of the smaller solenoid are connected together. The smaller solenoid is then removed from inside the larger solenoid.

With reference to magnetic fields, explain why a force is needed to remove the smaller solenoid.

Answer/Explanation

Ans:

(a) less in smaller solenoid

(b) greater in smaller solenoid

such as to (produce effects that) oppose the change that caused it

(ii) change of flux (linkage) in smaller solenoid induces e.m.f. in smaller solenoid

(induced) current in smaller solenoid causes field around it

the two fields (interact to) create an attractive force

Question

(a) Alternating current (a.c.) is converted into direct current (d.c.) using a full-wave rectification circuit. Part of the diagram of this circuit is shown in Fig. 7.1.

(i) Complete the circuit in Fig. 7.1 by adding the necessary components in the gaps.

(ii) On Fig. 7.1 mark with a + the positive output terminal of the rectifier.

(b) The output voltage V of an a.c. power supply varies sinusoidally with time t as shown in Fig. 7.2.

(i) Determine the equation for V in terms of t, where V is in volts and t is in seconds.

(ii) The supply is connected to a \(12\) Ω resistor. Calculate the mean power dissipated in the resistor.

Answer/Explanation

Ans:

(a) (i) two diodes added in correct directions (Both diodes pointing inwards and upwards), correct symbols only

(ii) ‘+’ anywhere on upper output wire

(b) (i) \(ω\) = \(2π / T\)

= \(2π / 2.5\)

= \(0.80 π\) or \(4 π / 5\) or \(2.5\)

\((V =)\) \(3.5 \sin (0.8 π t)\) or \(3.5 \sin (4 π t / 5)\) or \(3.5 \sin (2.5 t)\)

(ii) \((P =) \frac{V^{2}}{2R}\) or \((P=) \frac{V_{r.m.s^{2}}}{R}\)

= \(\frac{3.5^{2}}{2\times 12}\) or \(\frac{2.47^{2}}{12}\)

= \(0.51\) W

Question

(a) State the formula for the de Broglie wavelength λ of a moving particle.

State the meaning of any other symbol used.

(b) Electrons accelerate through a potential difference, pass through a thin crystal and are then incident on a fluorescent screen.

The pattern in Fig. 8.1 is observed on the fluorescent screen.

(i) State the name of the phenomenon shown by the electrons at the crystal.

(ii) State what this phenomenon shows about the nature of electrons.

(iii) Suggest why the thin crystal causes the phenomenon in (b)(i).

(iv) The electron is accelerated through a different potential difference. The new pattern observed on the screen is shown in Fig. 8.2.

State and explain the change that has been made to the potential difference to create the pattern shown in Fig. 8.2.

Answer/Explanation

Ans:

(a) \(\lambda = \frac{h}{p}\) or \(\lambda = \frac{h}{mv}\)

where h is the Planck constant and

p is the momentum (of particle) / mv is the momentum (of particle) / m is the mass (of particle) and v is the velocity (of particle)

(b) (i) (electron) diffraction

(ii) moving electrons behave like waves

(iii) spacing between atoms ≈ wavelength of electron

diameter of atom ≈ wavelength of electron

(iv) Any one of:

• wavelength has decreased

• electron had greater momentum

so (accelerating) p.d. was increased

Question

Polonium-\(211\) \(\left ( _{84}^{211}\textrm{Po} \right )\) decays by alpha emission to form a stable isotope of lead (Pb).

(a) Complete the equation for this decay.

(b) The variation with time t of the number of unstable nuclei N in a sample of polonium-\(211\) is shown in Fig. 9.1.

At time t = 0, the sample contains only polonium-\(211\).

(i) Use Fig. 9.1 to determine the decay constant λ of polonium-\(211\). Give a unit with your answer.

(ii) Use your answer in (b)(i) to calculate the activity at time t = 0 of the sample of polonium-\(211\).

(iii) On Fig. 9.1, sketch a line to show the variation with t of the number of lead nuclei in the sample.

(i) Calculate, in J, the total amount of energy given to alpha particles that are emitted between time \(t = 0.30\) s and time \(t = 0.90\)s.

(ii) Suggest why the total amount of energy released by the decay process between time t = \(0.30\) s and time t = \(0.90\) s is greater than your answer in (c)(i).

Answer/Explanation

Ans:

(a) \(207\), \(82\) for lead

\(4\), \(2\) for alpha

(b) (i) (half-life found as) \(0.52\) s or correctly read points substituted into

\(N\) = \(N_{0}e^{-2t}\)

\(\lambda\) = \(\frac{0.693}{t_{1/2}}\)

\(\lambda\) = \(\frac{0.693}{0.52}\)

\(\lambda\) = \(1.3\, s^{-1}\)

(ii) \(A\) = λN

\(12\) = \(1.3 \times 24 \times 10^{12}\)

\(13\) = \(3.1 \times 10^{13}\) Bq

(iii) upwards curve of decreasing gradient starting from \((0,0)\)

passes through \((0.52, 12)\) and \((1.2, 18.8)\)

\(6900 \times 10^{3} \times 1.6 \times 10^{-19}\)

\((16 \times 10^{12} – 7.2 \times 10^{12}) \times 6900 \times 10^{3} \times 1.6 \times 10^{-19}\)

= \(9.7\) J

(ii) lead nuclei have kinetic energy

gamma photons are also emitted

Question

In an X-ray tube, electrons are accelerated through a potential difference of \(75\) kV. The electrons then strike a tungsten target of effective mass \(15\) g.

The electron energy is converted into the energy of X-ray photons with an efficiency of \(5.0\) %. The rest of the energy is converted into thermal energy.

(a) The X-ray tube produces an image using a current of \(0.40\) A for a time of \(20\) ms.

The specific heat capacity of tungsten is \(130\, J\, kg^{–1}\, K^{–1}\).

Determine the temperature rise ΔT of the tungsten target.

(b) The linear attenuation coefficient of the X-ray photons in muscle is \(0.22\, cm^{–1}\).

Calculate the thickness t of muscle that will absorb \(80\) % of the incident X-ray intensity.

Table 10.1

Two X-ray images are taken, one of equal thicknesses of bone and muscle and another of equal thicknesses of blood and muscle.

Explain why one of these images has good contrast, but the other does not.

Answer/Explanation

Ans:

(a) energy = mc ΔT

energy = ItV

\((\Delta T =)\frac{0.40 \times 0.020 \times 75\, 000 \times 0.95}{0.015 \times 130}\)

= \(290\) K

(b) \(I\) = \(I_{0}e^{-\mu t}\)

\(0.20\) = \(e^{-0.22t}\)

\(t\) = \(7.3\) cm

(linear) attenuation coefficients / μ very different for bone and muscle

(very) different amounts (of X-rays) absorbed so good contrast

or (very) different intensities transmitted so good contrast

(linear) attenuation coefficients / μ similar for blood and muscle

similar amounts (of X-rays) absorbed so poor contrast

or similar intensities transmitted so poor contrast

Question

Positron emission tomography (PET scanning) obtains diagnostic information from a person. The information is used to form an image.

(a) PET scanning uses a tracer.

Explain what is meant by a tracer.

(b) PET scanning involves annihilation.

(i) Explain what is meant by annihilation.

(ii) State the names of the particles involved in the annihilation process.

(ii) Calculate the wavelength of each gamma photon released.

(d) Explain how the gamma photons are used to produce an image.

Answer/Explanation

Ans:

(a) substance containing radioactive nuclei that is introduced into the body or

substance containing radioactive nuclei that is absorbed by the tissue being studied

(b) (i) a particle interacting with its antiparticle so that mass is converted into energy

(ii) electron(s) and positron(s)

\(2\times 9.11\times 10^{-31}\times 3.00\times 10^{-82}\)

= \(1.64 \times 10^{-13}\) J

(ii) \(\lambda\) = \(\frac{2hc}{E}\)

= \(\frac{2\times 6.63\times 10^{-34}\times 3.00\times 10^{8}}{1.64\times 10^{-13}}\)

= \(2.43 \times 10^{-12}\) m

(d) Any \(3\) from:

• the two gamma photons travel in opposite directions

• gamma photons detected (outside body / by detectors)

• gamma photons arrive (at detector) at different times

• determine location of production (of gamma)

• image of tracer concentration in tissue produced

Question

(a) State what is meant by luminosity of a star.

(b) The luminosity of the Sun is \(3.83 \times 10^{26}\) W. The distance between the Earth and the Sun is \(1.51 \times 10^{11} m\).

Calculate the radiant flux intensity F of the Sun at the Earth. Give a unit with your answer.

(d) The radius of the Sun is \(6.96 \times 10^{8}\) m.

Show that the temperature T of the surface of the Sun is \(5770\) K.

(e) The wavelength \(λ_{max}\) of light for which the maximum rate of emission occurs from the Sun is \(5.00 \times 10^{–7}\)m.

The temperature of the surface of the star Sirius is \(9940\) K.

Use information from (d) to determine the wavelength of light for which the maximum rate of emission occurs from Sirius.

Answer/Explanation

Ans:

(a) total power of radiation emitted (by the star)

(b) \(F\) = \(\frac{L}{4\pi d^{2}}\)

= \(\frac{3.83\times 10^{26}}{4\times \pi 1.51\times 10^{112}}\)

= \(1340 W\, m^{-2}\)

= \(\frac{3.83\times 10^{26}}{3.00\times 10^{82}}\)

= \(4.26 \times 10^{9}\) kg

(d) \(L = 4\pi \sigma r^{2}T^{4}\)

\(3.83 \times 10^{26}\) = \(4\times \pi 5.67\times 10^{-8}\times 6.96 \times 10^{82}\times T^{4} leading to T\) = \(5770\) K

(e) \(\lambda _{(max)} \frac{1}{T}\)

\(\frac{5.00 \times 10^{-7}}{\lambda} = \frac{9940}{5770}\)

\(\lambda = 2.90 \times 10^{-7}\) m

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